为什么不接受Scala的符号作为列引用? [英] Why is Scala's Symbol not accepted as a column reference?

问题描述

尝试使用Spark SQL的示例，它们看起来很不错，除非需要使用表达式:

Trying the examples of Spark SQL, they seem to work well except when expressions are needed:

scala> val teenagers = people.where('age >= 10).where('age <= 19).select('name)
<console>:23: error: value >= is not a member of Symbol
       val teenagers = people.where('age >= 10).where('age <= 19).select('name)

scala> val teenagers = people.select('name)
<console>:23: error: type mismatch;
 found   : Symbol
 required: org.apache.spark.sql.catalyst.expressions.Expression
       val teenagers = people.select('name)

似乎我需要一个未记录的导入.

It seems that I need an import not documented.

如果我批量导入所有内容

If I bulk importing everything

import org.apache.spark.sql.catalyst.analysis._
import org.apache.spark.sql.catalyst.dsl._
import org.apache.spark.sql.catalyst.errors._
import org.apache.spark.sql.catalyst.expressions._
import org.apache.spark.sql.catalyst.plans.logical._
import org.apache.spark.sql.catalyst.rules._ 
import org.apache.spark.sql.catalyst.types._
import org.apache.spark.sql.catalyst.util._
import org.apache.spark.sql.execution
import org.apache.spark.sql.hive._

...和

val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext._

有效.

推荐答案

您缺少一个隐式转换.

val sqlContext: org.apache.spark.sql.SQLContext = ???
import sqlContext._

但是，在最新(和受支持的)Spark版本中，情况已经改变.

That has however changed in the recent (and supported) versions of Spark.

这篇关于为什么不接受Scala的符号作为列引用?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！